I am currently writing a function for solving optimisation problems were the inputs are varing:
function solve_problem!(f, x0, lvar, uvar, c, lcon, ucon); #If there are constraints
function solve_problem!(f, x0, lvar, uvar); #If there are no constraints.
I have solved this problem by doing as follows;
function solve_problem!(f, x0, lvar, uvar, c = 0, lcon = 0, ucon = 0;)
if c !=0
nlp = ADNLPModel(f_mod, x0, lvar, uvar, c, lcon, ucon);
else
nlp = ADNLPModel(f_mod, x0, lvar, uvar)
end
end
However, is there a better way of doing this? It is working, but I don't feel like it is good coding. Thank you!
I can't think about another way of doing this as ADNLPModel is an allready existing function.
Why not to use the same approach as is already used by ADNLPModel?
Julia is a multiple dispatch language - you can and should have the same method name differentiated by the number of paramaters (and perhaps their types as well).
function solve_problem!(f, x0, lvar, uvar, c, lcon = 0, ucon = 0)
nlp = ADNLPModel(f_mod, x0, lvar, uvar, c, lcon, ucon)
do_something(nlp)
end
function solve_problem!(f, x0, lvar, uvar)
nlp = ADNLPModel(f_mod, x0, lvar, uvar)
do_something(nlp)
end
If you for some reason were not able to use multiple dispatch and still need to check for parameter assignment I would suggest using nothing as the default value rather than 0.
Related
Andrew Ng's course in Coursera, which Stanford's Machine Learning course, features programming assignments that deal with implementing the algorithms taught in class. The goal of this assignment is to implement linear regression through gradient descent with an input set of X, y, theta, alpha (learning rate), and number of iterations.
I implemented this solution in Octave, the prescribed language in the course.
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
numJ = size(theta, 1);
for iter = 1:num_iters
for i = 1:m
for j = 1:numJ
temp = theta(j) - alpha /m * X(i, j) * (((X * theta)(i, 1)) - y(i, 1));
theta(j) = temp
end
prediction = X * theta;
J_history(iter, 1) = computeCost(X,y,theta)
end
end
On the other hand, here is the cost function:
function J = computeCost(X, y, theta)
m = length(y);
J = 0;
prediction = X * theta;
error = (prediction - y).^2;
J = 1/(2 * m) .* sum(error);
end
This does not pass the submit() function. The submit() function simply validates the data through passing an unknown test case.
I have checked other questions on StackOverflow but I really don't get it. :)
Thank you very much!
Your gradient seems to be correct and as already pointed out in the answer given by #Kasinath P, it is likely that the problem is that the code is too slow. You just need to vectorize it. In Matlab/Octave, you usually need to avoid for loops (note that although you have parfor in Matlab, it is not yet available in octave). So it is always better, performance-wise, to write something like A*x instead of iterating over each row of A with a for loop. You can read about vectorization here.
If I understand correctly, X is a matrix of size m*numJ where m is the number of examples, and numJ is the number of features (or the dimension of the space where each point lies. In that case, you can rewrite your cost function as
(1/(2*m)) * (X*theta-y)'*(X*theta-y);%since ||v||_2^2=v'*v for any vector v in Euclidean space
Now, we know from basic matrix calculus that for any two vectors s and v that are functions from R^{num_J} to R^m, the Jacobian of s^{t}v is given by
s^{t}Jacobian(v)+v^{t}*Jacobian(s) %this Jacobian will have size 1*num_J.
Applying that to your cost function, we obtain
jacobian=(1/m)*(theta'*X'-y')*X;
So if you just replace
for i = 1:m
for j = 1:numJ
%%% theta(j) updates
end
end
with
%note that the gradient is the transpose of the Jacobian we've computed
theta-=alpha*(1/m)*X'*(X*theta-y)
you should see a great increase in performance.
your computecost code is correct
and Better follow the vectorized implementation of Gradient Descent.
You are just iterating and it is slow and may have error.
That course aims you to do vectorized implementation as it is simple and handy at the same time.
I knew this because I did that after sweating a lot.
Vectorization is good:)
I am doing a simple task on ocatve.
I have to administer drug dose at day 1,7,21 and 28..
i wrote a function like that:
function xdot = f (x,t)
a=[1;7;21;28]
drug=0; ### initially drug is zero
if (t==a)
drug=drug+57.947;
else
drug=drug+0;
endif
xdot=(-0.4077)*(x)+ drug; passing the value of drug to differential equation
endfunction
In the main file i called this function in lsode:
t=linspace(0,30,30);
x0=0;
y=lsode(#ex,x0,t); ### ex is the file name where function is written
plot(t,y,'o')
This program doesn't work.. it displays all the time zero value for drug. Can anybody help me that how to administer dug dose with certain time step by manipulating linspace function.
It looks like you have a simple clearance model, and at each time in a, you want the dose to be delivered instantaneously. That is, at each time in a, the amount of drug in the subject increases by 57.947.
If that is the model that you have in mind, implementing it in the formula for xdot will not work well. You would actually need to implement it as a "delta function", and lsode won't work with that.
Instead, I recommend solving the problem in stages, corresponding to the time intervals [0, 1], [1, 7], [7, 21], and [21, 28]. During one stage, the differental equation is simply xdot = -0.4077*x. In the first stage, the initial condition is 0. In the next stage, the initial condition is the final value of the previous stage plus the dosage amount 57.947.
Here's a script:
dose = 57.947;
a = [1 7 21 28 30];
x0 = 0.0;
t0 = 0;
t = [];
y = [];
for t1 = a
tinterval = linspace(t0, t1, 2*(t1 - t0) + 1);
yinterval = lsode(#(x, t) -0.4077*x, x0, tinterval);
t = [t tinterval];
y = [y yinterval'];
t0 = t1;
x0 = yinterval(end) + dose;
end
plot(t, y, "linewidth", 2);
The script creates this plot:
Note that the differential equation xdot = -k*x has the solution x0*exp(-k*(t-t0)), so the call to lsode could be replaced with
yinterval = x0*exp(-0.4077*(tinterval - t0));
If you do this, also remove the transpose from yinterval two lines below:
y = [y yinterval];
If you want to keep the implementation of the administration of the drug dose within the formula for xdot, you'll need to distribute it over a small time interval. It could be implemented as a short rectangular pulse of width w and height 57.974/w. You'll also need to ensure that lsode takes sufficiently small internal time steps (less than w) so that it "sees" the drug dose.
You probably want replace t==a by ismember (t, a). Also, you should erase the else clause as it has no effect on the answer.
UPDATE
Consider rewriting your function as:
function xdot = f (x,t)
xdot = -0.4077*x + 57.947*ismember (t, [1 7 21 28])
endfunction
I have some problems with integrating in Octave.
I have the following code:
a=3;
function y = f (x)
y = x*x*a;
endfunction
[v,ier,nfun,err]=quad("f",0,3);
That 'a' in function is giving me troubles.
Octave says that 'a' is undefined. So, if I instead of 'a' put number 3 in function y everything works just fine. However, I want to have 'a' in function so I can change it's value.. How do I do that?
Thanks
You could use a function closure, which will encapsulate the a.
function f = makefun (a)
f = #(x) x * x * a;
endfunction
f = makefun(3)
[v, ier, nfun, err] = quad(f, 0, 3);
There are two main options.
Option 1 is, as voithos notes, make 'a' an input to the function.
Option 2 is to define 'a' to be a global variable.
global a=3;
function y = f (x)
global a
y = x*x*a;
endfunction
[v,ier,nfun,err]=quad("f",0,3);
This will cause 'a' to be the same value inside and outside the function.
Your function is actually dependent on two values, x and a, therefor:
f=#(x,a) x*x*a
[V, IER, NFUN, ERR] = quad (#(x) f(x,3), A, B, TOL, SING)
I used inline functions as i think it is easier to understand.
I am learning MatLab on my own, and I have this assignment in my book which I don't quite understand. Basically I am writing a function that will calculate sine through the use of Taylor series. My code is as follows so far:
function y = sine_series(x,n);
%SINE_SERIES: computes sin(x) from series expansion
% x may be entered as a vector to allow for multiple calculations simultaneously
if n <= 0
error('Input must be positive')
end
j = length(x);
k = [1:n];
y = ones(j,1);
for i = 1:j
y(i) = sum((-1).^(k-1).*(x(i).^(2*k -1))./(factorial(2*k-1)));
end
The book is now asking me to include an optional output err which will calculate the difference between sin(x) and y. The book hints that I may use nargout to accomplish this, but there are no examples in the book on how to use this, and reading the MatLab help on the subject did not make my any wiser.
If anyone can please help me understand this, I would really appreciate it!
The call to nargout checks for the number of output arguments a function is called with. Depending on the size of nargout you can assign entries to the output argument varargout. For your code this would look like:
function [y varargout]= sine_series(x,n);
%SINE_SERIES: computes sin(x) from series expansion
% x may be entered as a vector to allow for multiple calculations simultaneously
if n <= 0
error('Input must be positive')
end
j = length(x);
k = [1:n];
y = ones(j,1);
for i = 1:j
y(i) = sum((-1).^(k-1).*(x(i).^(2*k -1))./(factorial(2*k-1)));
end
if nargout ==2
varargout{1} = sin(x)'-y;
end
Compare the output of
[y] = sine_series(rand(1,10),3)
and
[y err] = sine_series(rand(1,10),3)
to see the difference.
I'm new with Matlab. I hope you can help me.
I have to solve a system of ODEs using ODE45 function. Here is the function which describes my equitions.
function dNdt = rateEquations(t, y)
%populations of corresponding state
Ng = y(1);
Ns = y(2);
Nt = y(3);
%All constants used are dropped for the sake of easy reading.
Note the parameter F.
%rate equations
dNs = s0 * Ng * F - Ns/ t_S1;
dNt = Ns / t_ISC - Nt / t_T1;
dNg = -dNt - dNs;
dNdt = [dNg; dNs; dNt];
end
Then, in my script .m-file i call the ode45 function in 'for loop'. During each iteration i have to change the parameter F and pass it to my 'rateEquations' - function. But i don't know how to realize it.
for T = Tmin: dt : Tmax
%initial conditions
initialConditions = [N0 0 0];
timeSpan = [T T+dt];
before calling ODE45 F is to be changed.
[t,N] = ode45('rateEquations', timeSpan, initialConditions)
and so on ...
end
Thanks in advance.
You want make F an argument of your derivative function and pass the right anonymous function to ode45:
[t,N] = ode45(#(t,y) rateEquations(t,y,F), timeSpan, initialConditions)